[Math] Why is every continuously differentiable function with a uniform bounded derivative lipschitz continuous

Question

I only know how to prove this for functions on a convex set by using the mean value theorem, but is this also true for this general case when nothing is said about the domain of the function besides the fact that it is a subset of $\mathbb{R}^n$?

Answer 1

The function $f(x) = \operatorname{sgn} x$ defined on the domain $D = \mathbb{R} \setminus \{0\}$ is differentiable at every point of $D$ and the derivative is continuous and bounded (it's identically zero), but it is not Lipschitz.